Need help with algebra word problems....?

Answer the ones you can.

1) John receives a total of $1282 a year from two investments. He has invested 4000 in one account and $8200 in another acount that pays 1.5% more than the first account. What is the interest rate for each account?

2) Grace needs to get a loan for a house that she is going to purchase. She discovers that she must get two loans instead of one large loan. She borrows $30,000 from one institution and $48,000 from another institution whose lending rate is 2% more than the other. If her total monthly interest charge on the loans is $600, what interest rate does she pay on each loan?

3) Rebecca invests money in two accounts. One bank pays 8.5% on her investment of $1200 while ther other pays 9% on her investment of $2000. How long is the term of investment if the total interest is $70.50?

Thanks for the help.

Need help with algebra word problems....?credit cards

1282 = 4000*x + 8200*(x+.015)

1282 = 12200x + 123

1159=12200x

.095=x so the first account pays 9.5%, the other pays 11%

30,000*x + 48,000*(x+.02) = 7,200 (annual interest of 600x12)

78,000*x + 960 = 7200

78,000*x = 6,240

x = 12.5 so the first loan = 12.5%, the second is 14.5%

((1,200*.085)+(2,000*.09) )*x = 70.50

(102 + 180)*x = 70.50

282*x = 70.50

x=70.50/282 = .25 or 1/4 year = 3 months

Need help with algebra word problems....?

loan

#1: use the amount John invested in the first account as X (the variable). So the amount he invested for the 2nd accound it 1.015X right? Then 4000X+8200(1.015X)=1282. Then solve.

The rest are all like that, just use a variable for one investment and then add/subtract from that variable for the other.